package algorithm.problems.bit_manipulation;

/**
 * Created by gouthamvidyapradhan on 01/12/2017.
 * The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

 Now your job is to find the total Hamming distance between all pairs of the given numbers.

 Example:
 Input: 4, 14, 2

 Output: 6

 Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
 showing the four bits relevant in this case). So the answer will be:
 HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
 Note:
 Elements of the given array are in the range of 0 to 10^9
 Length of the array will not exceed 10^4.

 Solution: O(N * 32): Count the number of set bits in each of 32 bit positions and then take the sum of
 product of number of set bits x number of un-set bits
 */
public class TotalHammingDistance {
    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        int[] A = {1000000000, 4, 14, 2};
        System.out.println(new TotalHammingDistance().totalHammingDistance(A));
    }

    public int totalHammingDistance(int[] nums) {
        int sum = 0;
        for(int i = 0; i < 32; i ++){
            int numOfOnes = 0;
            int p = (1 << i);
            for (int num : nums) {
                if ((num & p) > 0) {
                    numOfOnes++;
                }
            }
            sum += ((nums.length - numOfOnes) * numOfOnes);
        }
        return sum;
    }
}
